Acid Base Titrations in Aqueous Solvents

Transcription

1 Acid Base Titrations in Aqueous Solvents Introduction: All kind of titrations in various samples are performed today in process analysers and laboratories, by far the most of them are acid base titrations. Acids and bases play an important role in many processes and are controlled at a regularly basis. In most cases titration in water is quite adequate to determine the acid or base concentration. Generally for the titration of acids a strong base like sodium or potassium hydroxide is used. For the titration of bases a very strong acid like hydrochloric acid is used, the hydrochloric acid can be replaced by nitric acid or sulphuric acid if the sample contains substances reacting with chloride such as silver ions. Titration is an absolute method so requires no calibration, on top of that multiparameter determinations are possible: with one acid as a titrant several bases can be determined. Titration is a quite versatile technique where it concerns the concentration range that can be analysed. Variations in sample size (between 0. and 0 ml), in concentration of the titrant (between 0.0 M and M) and that of the burette size (between ml and 0 ml) can be made. Most determinations in this monograph are potentiometric titrations with a glass electrode as indication system. In section an example is given with an antimony electrode and section on a photometric indication. ) Classification of Acids according to strength: An acid dissolved in water gives the next reaction: HA + H O H O + + A Depending on the strength of an acid the equilibrium lies to the right or to the left. A very strong acid, like hydrochloric acid completely dissociates in water. A weak acid, like acetic acid only partly dissociates and most of the acetic acid is present in the form of the non dissociated acid. For the classification of the acid strength the dissociation constant Ka is used. + [ H O ][ A ] K a = [ HA ] Example: Calculation of the part of the acid that has not been dissociated. An Acid HA with Ka= is dissolved in water at a concentration of mol / l Reaction: HA + H O H O + + A So [H O+] is equal to [A ] The concentration is mol / l in total, so [HA] + [A ] = mol / l + H O ][ A ] [ A ] K a = [ [ HA ] = [ HA The greater part of the total concentration HA is in non dissociated form HA so [HA] ~ mol/l and [A ] = and [A ] = So about % of the acid has been dissociated and % is not. ] Definition pka = log Ka For easy operations normally the unit pka is used for the acid strength. pka means the log of the Ka value. Example: Ka of Acetic Acid is:. * So log Ka =. and the pka =.

2 No. Acid Base Titrations in Aqueous Solvents a) Classification of Acids according to pka : Some examples of acids, besides the listed ones many more of them can be found in literature. Very Strong Acids pka < 0 pka HClO HI HCl H SO HNO Strong Acids 0 < pka<. HSO. H PO. HF. Formic Acid. Weak Acids. < pka <. Acetic Acid. H PO. HCN. Very Weak Acids pka >. HPO. Phenol The very strong acids all completely dissociate in water, the differences in pka value in this group are only noticeable in nonaqueous solvents such as acetone and alcohols. In water all the very strong act the same. The figure below presents titration curves of acids with different pk values using the degree of conversion λ versus. λ = 0 at start of the titration and λ = in the Endpoint Fig. Acids with pk numbers from.00 pk λ

3 No. Acid Base Titrations in Aqueous Solvents As can be seen from the curves: the potential jump gets smaller with higher pk values acids can be titrated with a reasonable accuracy till a pk of very weak acids show no inflection point at all! for the weaker acids (in the range of pk. and.) at the factor λ = 0. the pk values correspond to the values. b) Classification of Bases according to pk b : A base dissolved in water gives the next reaction: B + H O OH + HB + Strong bases are for instance NaOH, KOH, and tetra methyl ammonium hydroxide. Analogue to acids a strong base reacts completely and a weak base partly dissociates. The strength of a base is given by the K b value, following the formulae: + [ OH ][ HB ] K b = [ B] For easy operations the next equation is used: Definition pk b = log K b Some examples of bases Weak bases. < pkb <. Very weak bases pkb >. Ammonia. Caffeine. Pyridine. Acetamid. Aniline. Urea. Equation: pka+pkb=pkw Water dissociates for a very small part in [H O + ] and [OH ], following the reaction: H O H O + + OH, the K of this reaction Kw = and the pkw = Ammonia reacts with water: NH + H0 NH + + OH The (NH ) + reacts with water NH + + H0 NH + H O + Because the ratio between [H O + ] and [OH ] is known by the pkw, the pka and pkb can easily be calculated if one of both is known because pka+pkb = pkw =.0 So Ammonia (NH ) is a weak base with pkb =. Therefore the Ammonium (NH ) + is a weak acid with pka =. In literature many tables of the pka of acids and pkb of bases can be found, sometimes the constant is represented just as pk and the reader has to find out himself if is the stated pk is the pka or the pkb. Unfortunately some tables do not mention the pkb of bases but just record the pka of the corresponding acid. Because the sum of pka and pkb is, calculation of pkb of the corresponding base is easy. a) Polybasic Acids: Some acids react with water in steps; a wellknown example is Phosphoric Acid (Fig. a) During the titration H PO reacts with hydroxide in three steps. H PO + OH H PO + H O with pka =. H PO + OH HPO + H O with pka =. HPO + OH PO + H O with pka =. A titration could theoretically result in three endpoints. The last EP however cannot be detected because of the high pka number (fig a).

4 No. Acid Base Titrations in Aqueous Solvents Fig. a H PO with 0. M NaOH EP 0 Fig. b Na PO with 0. M HCl EP 0 b Polyacid bases: The Na PO sodium salt can be titrated with HCl ( fig. b), so again a titration in steps that ends in H PO. The pkb values can be calculated from the pka numbers of phosphoric acid, using the expression pka + pkb = pkw=. The corresponding pkb numbers are respectively.,. and.. Again the last EP cannot be detected because the base is too weak. Separation of a Mixture of Acids first idea Mixtures of two acids can be determined in one titration if there is a significant difference in pka value. The strongest acid is titrated first, resulting in the first endpoint, the second endpoint can represent the weaker acid. If the difference in pk is too small no separate endpoints will be found, only the sum of the acids can be determined. In this section some examples are given a) HCl and HNO The determination of hydrochloric acid and nitric acid is showed in fig.. The pka numbers are respectively for the hydrochloric and for the nitric acid. As mentioned earlier these negative values have no meaning in water and both acids act similar. The result is that no seperation occurs. The sum of both acids of coarse can easily be determined. In this particular case the HCl, containing chloride can be determined by a titration with AgNO. The difference between the results in the titration with NaOH and the result in the titration with AgNO represents the HNO. Fig. HCl (pk a ) and HNO (pk a ) HCl + HNO with 0. M NaOH No separation Sum can be determined

5 No. Acid Base Titrations in Aqueous Solvents b) HCl and Phosphoric Acid Solutions of hydrochloric acid are dissociated for 0 %, solutions of phosphoric acid are dissociated for about %. These values as such, indicate that a titration of a mix of both acids will give problems in detecting the first endpoint. And indeed no separation occurs. The first endpoint detected in the mix represents the sum in the HCl and first endpoint of H PO. Fortunately the endpoint from the reaction H PO + OH HPO + H O can be used for the determination of the phosphoric acid. Because the same consumption of hydroxide is used for the reaction H PO + OH H PO + H O the determination of HCl is possible too, just by subtracting the difference of the detected endpoints from the first endpoint. Fig. HCl ( pk a ) H PO ( pk a.) H PO ( pk a.) No separation H PO and HCl Separation H PO and H PO Titrant 0. M NaOH A) H PO B) HCl C) HCl+ H PO A B C c) Hydrochloric and Acetic Acid Both acids can be determined, the separation even occurs if the ratios hydrochloric acid and acetic acid are varied from to. In figure curves are shown of the ratios, and. The endpoint in this mixture of hydrochloric acid is about.. HCl alone gives an endpoint at about (see fig. ). The inflection of the hydrochloric acid in the mix is much smaller because of the presence of the inflection of the acetic acid. HCl (pk a ) Fig. HCl + Acetic Acid with 0. M NaOH Acetic Acid (pk a.) Separation HCl and Acetic acid A) HCl Acetic Acid B) C) EP EP EP A B C 0

6 No. Acid Base Titrations in Aqueous Solvents d) Hydrochloric Acid + Formic Acid The titration results in a poor inflection for the hydrochloric acid. Accuracy will be far from optimal, furthermore if the performance of the electrode declines false endpoints will be found. To check inflections, the differentiated ERC curve as represented in fig.b can be viewed. Large potential breaks in the normal curve, like the second endpoint, result in sharp peaks and the top reaches high ERC values above 0. The first endpoint is a poor inflection: the top reaches a value of. Into account has to be taken that this peak is on a sloping baseline, after correction the ERC value drops to. This figure is definitely too low for quantitative measurements. Discrete values of are the minimum, and advised is for a reasonable accuracy. So for mixed acids of this composition only the total of acids can be determined. Fig. a Fig. b HCl (pk a ) HC l + Formic Acid 0 HC l+formic Acid Formic Acid (pk a.) Too poor separation EP ERC EP e) Phosphoric acid and Acetic Acid The EP from the acetic acid is the smallest inflection. EP and have larger inflections, it is advised to use these endpoints for the calculation of the acetic acid. The phosphoric acid can be calculated from the. The difference between EP and represents the sum of the consumption of the H PO and the acetic acid. As for phosphoric acid solutions the consumption of the H PO and the H PO consumption are basically the same, the difference is the acetic acid. pka H PO. H PO. Acetic Acid. Poor seperation Fig. a Phosphoric+ Acetic Acid with 0. M NaOH EP EP Fig. b Phosphoric + Acetic Acid EP EP ERC 0 0 0

7 No. Acid Base Titrations in Aqueous Solvents f) HPO + Formic acid: No separation of the H PO and the formic acid, the first detected endpoint represents the sum of both acids. Because the H PO step is well separated, calculation of the phosphoric as well as the formic is possible. Note in the differentiated curve the small hill around ml. The peak height is so low that the endpoint from the HPO is not detected. pka H PO. H PO. Formic Acid. Fig. a H PO +Formic Acid EP ERC Fig. b H PO +Formic Acid EP No separation H PO and Formic Acid Separation of a mixed bases Analogue to acids, the simultaneous determination of bases can be done if there is a significant difference in pkb of the bases. Is the difference too small then only the sum of both bases can be determined. The stronger base, having the lowest pkb value is titrated first the one with the highest pkb is titrated last. a Amino ethoxy ethanol and hydroxyl amine. Amino ethoxy ethanol pkb:. Hydroxylamine: pkb:.0 The amines are well separated, so good quantitative determinations are possible. See the differentiated curves in fig. b. Fig. a Fig. b. Amines with 0. M H SO A: A) Amino Amino Ethoxy ethoxy ethanol Ethanol B) Hydroxylamine A B : Hydroxyl amine C) Mix C : Mix 0 0 Amines with 0. M HSO A B C A B C ERC C B C

8 No. Acid Base Titrations in Aqueous Solvents b) Carbonate, Bicarbonate and Hydroxide The Carbonic acid H CO can dissociate in two steps, first in the formation of Bicarbonate (HCO ) with a pka of. and the second to Carbonate (CO ) with pka.. The H CO is not stable in water and falls apart in CO and H O. The carbonate and bicarbonate are stable bases and can be titrated with acid. Using the expression pka + pkb = pkw = the pkb can be calculated. So the pkb of Carbonate is. and the pkb of bicarbonate is. In samples with Carbonate and Hydroxide in about equal concentrations EPs exist (Curve fig.) : NaOH + HCl NaCl + H O EP: Na CO + HCl NaHCO + NaCl EP: NaHCO + HCl CO + H O +NaCl The first EP cannot be used for quantitative measurements, but all concentrations can be calculated from the last two remaining endpoints because the consumption for the Na CO is equal to the consumption of NaHCO. With low concentrations carbonate (curve in fig.) two endpoints are detected. : sum of NaOH + Na CO and EP: NaHCO ) High CO = and high OH ) Low CO = and high OH Fig. Hydroxide and Carbonate with 0. M HCl EP EP EP ) Rules for seperation of acids and bases based on the pk For acids rules of thumb The negative values for the pka of very strong acids have a meaning only in organic solvents. For the rule of thumb in water, use for all of the very strong acids in these calculations the value pk = 0.As can be seen in the foregoing examples successful seperation based on the difference of pk values is achieved easier with weaker acids. Strong acids: Difference in pka at least pka > Weak acids: Difference in pka at least pka > For bases rules of thumb: Strong bases: Difference in pkb at least pkb > Weak bases: Difference in pkb at least pkb >

9 No. Acid Base Titrations in Aqueous Solvents ) Titration of very weak acids For Ammonia (NH ) the pkb is., using the formulae pka+pkb= the pka for the corresponding acid the Ammonium ion (NH)+ is.. The (NH)+ ion can be titrated with sodium hydroxide, because the pka value is above the accuracy will be limited. As a result the curve shows a small inflection. The results of different quantities of a standard solution of ammonium chloride result in a relative standard deviation of 0. %. Note that the endpoint itself varies with the concentration. Fig. Different concentrations (NH ) + NHCl with 0. M NaOH 0 Standard solution of ammoniumchloride, ml of standard + 0 ml water Titrant ml burette 0. M NaOH, DET ( MPD =, drift 0. /min) No. ml of sample EP in ml EP in NHCl in g/l Average..0 Relative Standard deviation 0. % ) Examples of the formation of a second compound Sometimes chemistry can help to solve a titration problem, some acids or bases form on addition of an auxiliary solution a compound that can be titrated easier. a) Ammonia and Formaldehyde An example is the reaction of ammonia with formaldehyde, adding an excess of formaldehyde converts ammonia to the hexamethylene tetramine. Reaction: NH + HCOH (CH ) N + H O The hexamethylene tetramine ( pkb =.) is a much weaker base then ammonia (pkb =.). So for the titration of ammonia with acid, the addition of formaldehyde would not improve the titration. But if we add formaldehyde to ammonium chloride, the weak acid ammonium (pka.) is converted to a to the stronger hexamethylene tetramine ion (pka.). This improves the curves remarkable. To check the improvement of the formaldehyde addition some titrations were performed with NaOH 0. M and standard samples of HCl, NHCl and a mix of both (with the

10 No. Acid Base Titrations in Aqueous Solvents addition of formaldehyde ml % pre neutralised to ). When the determination of the ammonium is done in two determinations, the overshoot of titrant in the HCl titration has to be added to the consumption in the ammonium titration. NH Cl HCl mix in one titration ) HCl + NH Cl mix in two titrations, titration of the HCl ) HCl + NH Cl mix in two titrations, second titration of the NH Cl after addition of formaldehyde Fig. a Fig. b HCl, NHCl,HCl+NHCl EP HCl, ml Formaldehyde %, NHCl No formaldehyde addition Formaldehyde addition Sample HCl det. NH Cl det. Sample HCl () NH Cl (). NH Cl. ml HCl + NH Cl. ml. ml. ml HCl + NH Cl. ml.0 ml HCl + NH Cl. ml. ml HCl + NH Cl. ml. ml average. ml. ml Relative Stand. Dev 0.0 % 0. % Comparing the standard deviations it is obvious that the deviation of the NH Cl is relatively high compared to the HCl deviation. This example follows the rule that deviations will increase with increasing pk value. b) Boric acid and Mannitol On the addition of the sugar Mannitol the Boric Acid with a pka of. can be converted to the boric ester. Because this ester has a pk of titration results will improve remarkably....0 Fig. Boric acid Boric Acid and Mannitol () ().. 0

11 No. Acid Base Titrations in Aqueous Solvents ) Hydrogen Fluoride with an Antimony Electrode For potentiometric acid base titrations the glass electrode is used almost exclusively as indication system. As the glass is attacked by fluoride solutions below the of, the the alternative of an Antimony Electrode can be used here. Fig. HF with 0. M NaO H (Antim ony electrode) E P 0 V [m l] ) Acid Base Titrations using Color Indicators. One of the oldest forms of titration is the addition of color indicators for Endpoint indication. A color indicator is an organic dye that has different colors depending on the of the solution. An example is methyl orange, above. the color is yellow and below. the color is red. The region between. and. is called, the interval, between these two values the color change is from red to orange to yellow. For an acid having its endpoint at about. methylorange can be used. The titration can be performed visually, the addition of titrant is continued until the original red color changes to orange. Fig.. yellow red EP ml indicator interval Color change Methylorange.. red yellow Methyl red.. red yellow p Nitrophenol.. colorless yellow Phenolphthalein.. colorless red violet

12 No. Acid Base Titrations in Aqueous Solvents Thymolphtalein.. colorless blue The disadvantage of visual indication is when the titration curve has not a clear inflection, the color change of the indicator is to gradually then. Adding more indicator to the solution will introduce a second problem because the indicator itself is also an acid or base and so is consuming acid or base. When two acids or bases are present in the sample, for each determination a different indicator has to be used. In fig. a Sodium carbonate is titrated a fotometric indication using the indicators phenolphthalein (PP) and methylorange (MO.) is used. At the beginning of the titration the solution is red/violet and due to the color of the PP and the transmission is low. Adding acid will lower the and the PP will become colorless. At this stage the resulting solution gets yellow due to the yellow color of the MO and this results in a high Transmission. Continuing the titration the MO will change its color from yellow to red and the Transmission will drop again. The raise and fall in the transmission results in the detection of an endpoint. As a comparison fig. b gives the determination with a glass electrode for detection of the endpoint. Fig. a Fig. b Na CO, fotometric. 0 nm phenolphthaleine + methylorange addition Na CO with glass electrode, % T EP EP 0 0 Jaap Guijt 0 February 00